Let the vertex of an angle $ \mathrm{ABC} $ be located outside a circle and let the sides of the angle intersect equal chords $ \mathrm{AD} $ and $ \mathrm{CE} $ with the circle. Prove that $ \angle \mathrm{ABC} $ is equal to half the difference of the angles subtended by the chords $ A C $ and $ D E $ at the centre.

Given:

Let the vertex of an angle \( \mathrm{ABC} \) be located outside a circle and let the sides of the angle intersect equal chords \( \mathrm{AD} \) and \( \mathrm{CE} \) with the circle.

To do:

We have to prove that \( \angle \mathrm{ABC} \) is equal to half the difference of the angles subtended by the chords \( A C \) and \( D E \) at the centre.

Solution:

$AD = CE$

We know that,

An exterior angle of a triangle is equal to the sum of interior opposite angles.

This implies, in $\triangle BAE$,

$\angle DAE = \angle ABC+\angle AEC$........(i)

$DE$ subtends $\angle DOE$ at the centre and $\angle DAE$ in the remaining part of the circle.

This implies,

$\angle DAE = \frac{1}{2}\angle DOE$.........(ii)

Similarly,

$\angle AEC = \frac{1}{2}\angle AOC$..........(iii)

From (i), (ii) and (iii), we get,

$\frac{1}{2}\angle DOE = \angle ABC+\frac{1}{2}\angle AOC$

$\angle ABC =\frac{1}{2}(\angle DOE-\angle AOC)$ 

Hence proved.

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Updated on: 10-Oct-2022

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